Problem: Consider the parametric curve: $\begin{aligned} x&=\cos(2t) \\\\ y&=6t^{3} \end{aligned}$ Which integral gives the arc length of the curve over the interval from $t=a$ to $t=b$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\int_{a}^{b} \sqrt{4\sin^2(2t)+324t^4}\,dt$ (Choice B) B $\int_{a}^{b} \sqrt{\cos^2(2t)+36t^6}\,dt$ (Choice C) C $\int_{a}^{b} \sqrt{\cos^2(2t)-36t^6}\,dt$ (Choice D) D $\int_{a}^{b} \sqrt{4\sin^2(2t)-324t^4}\,dt$
Explanation: This is the formula for the arc length of a parametric curve over the interval $[a, b]$ : $\int_a^b\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\, dt$ [Where does this formula come from?] Let's find $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ : $\begin{aligned} \dfrac{dx}{dt}&=\dfrac{d}{dt}\left[\cos(2t)\right] \\\\ &=-\sin(2t)(2) \\\\ &=-2\sin(2t) \\\\\\ \dfrac{dy}{dt}&=\dfrac{d}{dt}\left[6t^{3}\right] \\\\ &=18t^2 \end{aligned}$ Now we can find the expression for the arc length: $\begin{aligned} &\phantom{=}\int_{a}^{b} \sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt \\\\ &=\int_{a}^{b} \sqrt{\left(-2\sin(2t)\right)^2+\left(18t^2\right)^2}\,dt \\\\ &=\int_{a}^{b} \sqrt{4\sin^2(2t)+324t^4}\,dt \end{aligned}$ In conclusion, this integral gives the arc length of the curve over the interval from $t=a$ to $t=b$ : $\int_{a}^{b} \sqrt{4\sin^2(2t)+324t^4}\,dt$